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We wish to solve
Note that the solution coincides with the infinities of . As f(x) is linear, it has has one real, which we shall call . Furthermore, the root must be between -1 and 1, since f(1) = 12 > 0, and f(-1) = -20 < 0. Hence the unit-circle contour bounds a singularity in 1/f(x). We can therefore write
by the residue theorem.
Now the integral can be explicitly evaluated by making the substitution and hence . Therefore
This integral may be evaluated by way of an infinite series:
It can be easily seen that only the first term in the series contributes, leaving us with
Therefore, we have
Now, it can be shown that the residue of the reciprocal of a linear function of the form f(z) = az-b = 0 at f(z)=0 is 1/a, hence we have that
Where if and otherwise. Now, finally, we have
simplified, we have
and hence .
